A NOTE ON THE JORDAN CANONICAL FORM

A NOTE ON THE JORDAN CANONICAL FORM H. Azad Department of Mathematics and Statistics King Fahd University of Petroleum & Minerals Dhahran, Saudi Arabia hassanaz@kfupm.edu.sa Abstract A proof of the Jordan canonical form, suitable for a first course in linear algebra, is given. The proof includes the uniqueness of the number and sizes of the Jordan blocks. The value of the customary procedure for finding the block generators is also questioned. 2 MSC: 15A21. The Jordan form of linear transformations is an exceeding useful result in all theoretical considerations regarding conjugacy classes of matrices, nilpotent orbits and the Jacobson -Morozov theorem. The author wishes to share a proof of the Jordan form which he found in connection with a problem in Lie theory. The ideas of the proof give at the same time the number and sizes of all the blocks. The proof has the added advantage that the most important parts can be taught in a first course on linear algebra, as soon as basic ideas have been introduced and the invariance of dimensions has been established. It is thus also a contribution to the teaching of these ideas. Although extensive work has been done in [4] regarding this circle of ideas, our method provides a very simple algorithm whose importance is shown through some simple examples. A classical reference for this topic is Smirnov s book [2,p.245-254]. There is a very well known proof due to Fillipov [1], which is also given in Strang s book [3, p. 422-425]. There is also a proof given in the Wikipedia [5]. The proofs in [3 & 5] do not give sufficient details regarding the number of Jordan blocks and their sizes, nor an algorithmic procedure to handle matrices of large size. In view of the algorithm 1

given in this note, and the examples given below, it is not clear to us why a precise determination of the block generators is needed, although, for the sake of completeness, we have discussed this aspect too- at the expense of increasing the level of exposition. It would be very desirable to compare the computational complexity of computing the Jordan canonical form, using the algorithm given in this paper, with other algorithms, which programmes like Maple and Mathematica use to determine the Jordan form. As is well known, the main technical step in establishing the Jordan canonical form is to prove its existence and uniqueness for nilpotent transformations. We will return to the general case towards the end of this note. Let A be a nilpotent transformation on a finite dimensional vector space V, let v be a nonzero vector in V and n the smallest integer such that A n v =. Proposition 1 The vectors {A i v : i < n} are linearly independent. Proof. Take an expression n 1 c i A i v =, ( ) i= in which the number of non-zero coefficients is as small as possible. If the coefficient c j is the non-zero coefficient of largest index j, then multiplying by A n j, we obtain an expression like (*) of smaller length. So in (*) every c i with i < j is. Therefore c j A j v = and therefore A j v =, with j n 1, which contradicts the choice of n. This proves the claim, Proposition 2 Let R(A) be the range space of A and N(A) be the null space of A. Let {A(v i ) : i = 1,..., r} be a basis of the range space. Let {n j : j = 1,..., s} be a basis of the null space of A. Then {v i : i = 1,..., r, n j : j = 1,..., s} is a basis of the vector space V. Proof. Let v V. So A(v) = r c i A(v i ). Therefore v r c i v i belongs to the null space of A, hence it is a linear combination of the {v i } and {n j }. To see that these vectors are linearly independent, suppose r c i v i + s d j n j =. This gives r ca(v i ) = and by linear independence of the vectors A(v i ), we get c i =, i = 1,..., r. The linear 2 j=1

independence of {n j } then shows that d j =, j = 1,..., s. Proposition 3 V is a direct sum of cyclic subspaces. Proof. We prove this, as in the standard proofs [2,3], by induction on dimension. The null space of A is a non-zero subspace and therefore the range space of A is a proper subspace of V. If this is the zero subspace, then a basis of V gives the decomposition into cyclic subspaces. So suppose that R(A) is a nonzero subspace. It is an A invariant subspace. By induction on dimensions, it is a direct sum of cyclic subspaces, with generators v i, i = 1,..., k, and basis A j v i, j n i, and A ni+1 v i =. Let v i = Aw i. So A j v i = AA j w i shows, using Proposition 2, that the vectors A j w i, j n i are linearly independent. Also A ni+1 v i = A ni+2 w i =, so A ni+1 w i = A n i v i belong to the null space of A. By Proposition 2, if we enlarge A n i v i, i = 1,..., k, to a basis of the null space of A by adjoining independent vectors n 1,..., n l in the null space of A, then A j w i, j n i, i k, A n i v i,i = 1,..., k, n 1,..., n l form a basis of V. Therefore, the cyclic subspaces generated by w i, i = 1,..., k and the one-dimensional subspaces generated by n r, 1 r l give a direct sum decomposition of V into cyclic subspaces. From this description, it is clear that in each summand only A n i+1 w i = A n i v i contributes to the null space of A in that summand and therefore the number of summands in the above given decomposition is the dimension of the null space of A. Corollary Let d i = dim (N(A R(A i )), i =, 1,..., n), where n is the smallest positive integer so that A n =. The differences d d 1, d 1 d 2,..., d n 1 d n give the number of Jordan blocks of sizes 1, 2,..., n. Proof. As shown in the proof of Proposition 3, the number of summands in the Jordan decomposition is the dimension of the null space of A. Therefore the number of blocks of size 1 is dim(n(a)). Applying A removes all blocks, if any, of size 1, and so the number of blocks of size 2 is dim(n(a R(A))) = d 1. Continuing, we get that d i is the number of blocks of size i, i = 1,..., n. Therefore the difference d i 1 d i gives 3

the number of blocks of size i, for i = 1,..., n. Example Let A be any nilpotent upper triangular matrix whose entries to the right of the main diagonal give a non-singular matrix. Then the null space of A is 1 dimensional and therefore the canonical form of A consists of only one block. and In particular, the matrices 2 1 1 2 1 2 3 4 7 6 5 8 9 1 are conjugate matrices as they are conjugate to 1 1 1 1. In view of such examples, it is not clear to us why an algorithmic procedure is needed to find the precise generators of the various blocks, because all one needs to find the form of the Jordan blocks is to compute the invariants d i. Nevertheless, for the sake of completeness, we outline such a procedure - at the expense of increase in level of exposition. Step 1: Find all eigenvalues. For an eigenvalue λ, compute the generalized eigenspace corresponding to λ. Although, one needs to compute only all vectors annihilated by (A λi) dim V, it is algorithmically better to compute the vectors annihilated by (A λi) n, where n is the multiplicity of the eigenvalue λ in the characteristic polynomial of A. So, by working in the generalized eigenspace for λ, and replacing (A λi) by A, we may assume that A is a nilpotent transformation of index n. 4

So we assume now that A is a nilpotent transformation defined on a vector space V. Step 2: Find the number and sizes of blocks of this nilpotent transformation according to the algorithm given in the Corollary. This is the most important step, which is needed to complete the next step efficiently. Step 3: If, for some reason, one needs to find explicitly the generators of these blocks in terms of a preassigned basis, one can proceed as follows: For a vector v, call the smallest integer m so that A m (v) = the weight of v. There must be a vector of weight n; in fact, if we start with any basis of V, there must be such a vector in this basis. Call it v 1. The vectors v 1,..., A n 1 v 1 are then linearly independent by Proposition 1. Find a basis of N(A n )/N(A n 1 ). If v, w,... are representatives, then they are of weight n and are linearly independent. If x is in the span of v, Av,..., A n 1 v and A j x =, then x is a linear combination of A k v with k n j. It follows that v, Av,..., A n 1 v, w, Aw,..., A n 1 w, are linearly independent. Repeating this, we get a subspace W 1 generated by linearly independent vectors of weight n, and the number of such vectors is the number of blocks of size n. This takes care of blocks of size n. Consider the blocks of size m -just below n. Find a basis of N(A m )/(N(A m 1 ) W 1 ). Each basis of this generates a block of size m. Call this subspace W 2. Then W 1 + W 2 is a direct sum. Consider the blocks of size l just below m. Find a basis of N(A l )/(N(A l 1 ) (W 1 + W 2 )). Each basis element generates blocks of size l. Continuing in this way, we get all the blocks of the Jordan decomposition. Examples 5

1. If A = 1 1 1 1 then N(A)works out to be 1 dimensional, so there is only 1 Jordan block. Also so A 3 = N(A 3 ) = 1 and, as A 4 =, a basis of N(A 4 )/N(A 3 ) is ν =. 1 Therefore, this must be a generator of the block. x y z,, 2. Let A = 2 2 1 2 1 1 2 2 4 The eigenvalue 2 is of multiplicity 3, so the generalized eigenspace V (2) is 3 dimensional, whose basis works out to be (1,,, ), (, 1,, ), (,, 1, ) and the matrix of A V (2) is therefore We have 2 2 2 1 2 (A 2I) V (2) =.. 2 1 Let à = (A 2I) V (2). This gives d = dim(n(ã) = 2, d 1 = dim(n(ã R(Ã)) = 1, d 2 = dim(n(ã R(Ã2 )) =. Therefore à has d d 1 = 1 block of size 1 and d 1 d 2 = 1 block of size 2.. 6

The Jordan form of à is therefore 1 and of A V (2) is 2 2 1 2 The eigenspace for eigenvalue 4 is one-dimensional. Therefore, the Jordan form of A is 2 2 1 2 4 A final remark on applications: A main application of the Jordan form in differential equations is in computation of matrix exponentials. However, it is computationally more efficient to calculate the matrix of A relative to a basis of generalized eigenvectorsnot necessarily given by cyclic vectors -and compute its exponential relative to this basis; finally, conjugating by the change of basis matrix gives the exponential of A. Acknowledgments The author thanks the King Fahd University of Petroleum and. Minerals for its support and excellent research facilities. References [1] A. F. Filippov, A short proof of reduction to Jordan Form. Moscow Univ. Math. Bull. 236 (1971), 7-71. [2] V. I. Smirnov, Linear Algebra and Group Theory. (Dover Publications, New York, 1961). [3] Gilbert Strang, Linear Algebra and its Applications. (4th edition, Thomsons, Brooks/Cole, 26). [4] Steven Weintraub, Jordan Canonical Form: Theory and Practice. (Morgan and Claypool, 21). [5] http://en.wikipedia.org/wiki/jordan normal form 7